Question

4 dice are rolled. Probability of not getting same number on 3 dice

Answer 1

Please see the attachment

Answer 2

Answer:

167/177.

Step-by-step explanation:

Now, there are a total of 5 ways we can have the results.

i) All the 4 results are identical.    

Now, there are only 6 ways it can happen: 1111 , 2222 , 3333 , 4444 , 5555 , 6666.

ii) All the 4 results are different.  

Spaces for results: _  _  _  _

Possibilities          : 6  5  4  3

That means there are 6 possibilities for the first result [1,2,3,4,5,6]

Now we already got one result, for the second result that can't be got again(look at the condition in bold). Thus, for the second spot, there are only 5 possibilities.

Same applies for the 3rd spot- 4 possibilities.

For the last spot there are 3 possibilities.

So total number of ways possible to get result under this condition = (6 X 5 X 4 X 3) = 360

iii) Any 3 of the results are same

Spaces for results: _  _  _  _

Possibilities          : 6   1   1  5

Let's assume the first 3 results are identical.

So for the first spot, there can be 6 possible results [1 to 6]

But once put, 2nd and 3rd spot will be the same result, thus number of possibilities are 1.

For the last spot, it can be anyone except the result of the first 3, thus there are 5 possibilities for this spot.

So total number of ways possible to get result under this condition = (6 X 1 X 1 X 5) = 30

iv) Any 2 of the results are same and the other 2 are different from each other.

Spaces for results: _  _  _  _

Possibilities          : 6   1   5  4

The first result is got- 6 possibilities [1 to 6]

If we consider the first two spot to be identical, there are only one possibility for the 2nd spot.

For the 3rd spot, there can be 5 results and for the 4th spot there are 4 possibilities(check the condition, last two spots have to be different).

So total number of ways possible to get result under this condition = (6 X 1 X 5 X 4) = 120

v) Any 2 are same, as well as the other 2 are also same but 4 are not identical.

This is a tricky condition.

Spaces for results: _  _  _  _

Possibilities          : 6   1   5  1

Let's assume the first two results are same and the last two results are same, though both results differ from each other.

Example: 1122 , 2233 , 4455 etc.

For the first spot obviously there are 6 possibilities [1 to 6]

2nd result is thus determined and has only one possibility.

For the 3rd spot there are 5 possibility(except the result we already got at 1st and 2nd spot).

And as the 3rd and 4th place are identical, 4th spot has only one possibility.

So total number of ways possible to get result under this condition = (6 X 1 X 5 X 1) = 30

NOW, WE'VE MADE A MISTAKE IN THIS CONDITION. AS WE'VE COUNTED 1122 and 2211 as two different results, but think keenly, these two are same results.

So,  So total number of ways possible to get result under this condition= 15.

Therefore, number of possible outcomes = (6 + 360 + 30 + 120 + 15) = 531

Outcomes without the possibility to get any results with 3 identical numbers = (531-30) = 501

Thus, probability of not getting same number on 3 dices = (501/531) = 167/177